Troubles with a Dynamics exercise

Two carts (1&2) on a flat surface, are pushed by an external force (##\vec{F}##), exerted on 1 (the carts are motionless and touching each other).

Consider the two objects as particles and take no notice of any friction.

F=12N; mass of 1 (##m_1##)=4,0 kg; mass of 2 (##m_2##)= 2,0 kg.
Find the intensity and the direction of the force exerted by 1 on 2 (##\vec{F_{12}}##) and the force exerted by 2 on 1 (##\vec{F_{21}}##)

3. The attempt at a solution
I tried solving the system given by:
##\vec{F_{12}}=\vec{F} - \vec{F_{21}}## and ##\vec{F_{21}}= m_2 * a##

obtaining:
##m_1 * a = 12 - m_2 *a## ##\Rightarrow## ##a=2,0 m/s^2##

and thus:
##\vec{F_{21}}=2,0kg * (-2,0 m/s^2)=-4 N## with the minus sign, as this force is opposite to F
##\Rightarrow## ##\vec{F_{12}}=16N##
which, according to my textbook is not the right result.
I don't get where are the mistakes, though. Can anyone help me please?

Staff: Mentor

allright, should it be like this then?
i find the acceleration, which is:
##a=\frac{\vec{F}}{m_1+m_2}## = ##2 m/s^2##
in the free-body diagram of 1 there is ##\vec{F_{21}}##
so I multiply the acceleration for ##m_2##, which gives ##\vec{F_{21}}=-4N## (because its direction is opposite to that of the x-axis)
and, for Newton 3, there must be an equal and opposite force, which means ##\vec{F_{12}}=4N##

allright, should it be like this then?
i find the acceleration, which is:
##a=\frac{\vec{F}}{m_1+m_2}## = ##2 m/s^2##
in the free-body diagram of 1 there is ##\vec{F_{21}}##
so I multiply the acceleration for ##m_2##, which gives ##\vec{F_{21}}=-4N## (because its direction is opposite to that of the x-axis)
and, for Newton 3, there must be an equal and opposite force, which means ##\vec{F_{12}}=4N##

Yes, this looks right :)
You can check the answer by seeing that the resultant force on m1 is therefore 8N which agrees with its acceleration.